Calculus 1

The curve y = |x|/(sqrt(5- x^2)) is called a bullet-nose curve. Find an equation of the tangent line to this curve at the point (2, 2)

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1. Here is what your curve looks like
http://www.wolframalpha.com/input/?i=plot+y+%3D+%7Cx%7C%2F%28sqrt%285-+x%5E2%29%29

notice that -√5 < x < √5 is our domain, with x = ±√5 as asymptotes
Since (2,2) lies in quadrant I
I will use
y = x/√(5 - x^2) = x(5-x^2)^(-1/2)
dy/dx = x(-1/2)(5-x^2)^(-3/2) (-2x) + (5-x^2)^(-1/2)
at (2,2)
dy/dx = 2(-1/2)(1^(-3/2) (-4) + 1
= 4+1 = 5

so y = 5x + b
at (2,2)
2 = 10 + b
b = -8

tangent equation: y = 5x - 8

verification:
http://www.wolframalpha.com/input/?i=plot+y+%3D+%7Cx%7C%2F%28sqrt%285-+x%5E2%29%29+%2C+y+%3D+5x-8+from+1+to+2.2

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2. well, if x>=0, |x|=x, so we have

y = x/√(5-x^2)
y' = 5/(5-x^2)^(3/2)
At x=2, y'=5
So, the line through (2,2) with slope=5 is

y-2 = 5(x-2)

See this at

http://www.wolframalpha.com/input/?i=plot+y%3D|x|%2F%E2%88%9A%285-+x^2%29%2C+y%3D5%28x-2%29%2B2+for+0%3C%3Dx%3C%3D3

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